Point-Line Distance, in Coordinates

Let us recall some basic facts about Cartesian coordinates before going into the main topic of this post.

1. Any line of the form $Ax+By+C=0$ can be written as $y = -\frac{A}{B} x - \frac{C}{B}$. This implies that $-\frac{A}{B}$ is the slope of the line.

2. Perpendicular lines' slopes multiply to negative 1.

3. The distance from $(x_1,y_1)$ to $(x_2,y_2)$ is $\sqrt{(x_1-x_2)^2 + (y_1-y_2)^2}$.

We will use these to prove/compute numerous things.

Firstly, we will find the formula for the coordinates of the projection of a point $P$ or $(x_1,y_1)$ on to the line $l$ or $Ax+By+C=0$.

We know the $l$ has slope $-\frac{A}{B}$. Thus, any line perpendicular to it must have slope $\frac{B}{A}$. These are precisely facts (1) and (2) at the start of this post.

Now, we want to find a perpendicular to $l$ through $P$. It must have the form

$y = \frac{B}{A} x + n$ for some constant $n$. Plug in $(x,y) = (x_1,y_1)$ to get

$y_1 = \frac{B}{A} x_1+n$ or

$n = y_1 - \frac{B}{A} x_1$. Thus, we see that the perpendicular to $l$ through $P$ is has the above equation.

Now, the foot of the perpendicular from $P$ is clearly the intersection of $l$ and the above line. So we wish to find the common solution of

$y = \frac{B}{A} x + y_1 - \frac{B}{A} x_1$ and

$y = -\frac{A}{B} x - \frac{C}{B}$.

This looks horrible, but we are only solving for $x,y$. Everything else is just fancy butter that can be ignored! Subtracting yields

$0 = (\frac{B}{A} + \frac{A}{B} ) x + y_1 + \frac{C}{B} - \frac{B}{A} x_1$ or

$x = \dfrac{\frac{B}{A} x_1 - \frac{C}{B} - y_1}{\frac{A}{B} + \frac{B}{A}}$. Then further simplifying,

$x = \frac{B^2 x_1 - AC - ABy_1}{A^2+B^2}$

and (eww) we can substitute to see that $y$ equals

$\frac{A^2y_1-ABx_1-CB}{A^2+B^2}$.

Thus, we have proven a theorem.

Theorem: For any point $(x_1,y_1)$ not on the line $Ax+By+C=0$, the coordinates of the projection of $(x_1,y_1)$ onto the line are

$(\frac{B^2x_1-AC-ABy_1}{A^2+B^2}, \frac{A^2y_1-ABx_1-BC}{A^2+B^2})$.

Let us see an example.

Now we will do something greater- we will find the distance from the point $(x_1,y_1)$ to the line $l$.

Of course, the distance is just the length of the altitude from $P$ to $l$, so we need only compute, using the distance formula, the distance from

$(x_1,y_1)$ to $(\frac{B^2x_1-AC-ABy_1}{A^2+B^2}, \frac{A^2y_1-ABx_1-BC}{A^2+B^2})$.

Let's take this one step at a time- First, we know the expression equals using fact (3) from the beginning of this post.

$\sqrt{\left(\frac{x_1A^2+AC+ABy_1}{A^2+B^2}\right)^2 + \left(\frac{y_1B^2+ABx_1+BC}{A^2+B^2}\right)^2}$. We can simplify it as

$\frac{1}{A^2+B^2} \sqrt{(x_1A^2+AC+ABy_1)^2+(y_1B^2+ABx_1+BC)^2}$.

This is just ridiculous! But since we have a $A^2+B^2$ term already, we can kind of expect some $A^2+B^2$ to materialize from the expression inside the square roots to cancel it.

Indeed:

$\sqrt{(x_1A^2+AC+ABy_1)^2+(y_1B^2+ABx_1+BC)^2} = \sqrt{A^2 (x_1A+C+By_1)^2 + B^2 (y_1B+Ax_1+C)^2}$.

Thus, we see it equals

$\sqrt{(x_1A+y_1B+C)^2 (A^2+B^2)}$.

Be careful when taking square roots! Always remember the absolute value sign. Thus, our final expression is equal to

$\frac{|x_1A+y_1B+C|}{\sqrt{A^2+B^2}}$ and we have proven the second theorem.

** Theorem ** For any point $(x_1,y_1)$ and any line $Ax+By+C=0$ the distance from the point to the line equals

$\frac{|x_1A+y_1B+C|}{\sqrt{A^2+B^2}}$.

You should remember this second result as it is useful. For the first result, it will become natural to go through the process that we derived it with when you use it a lot.